Sindh Mcat Nts 2008 — Solved Past Paper with Answers

Q1. Destitute (Give Synonym)

• A. Disable
• B. Isolated
• C. Poor✓
• D. Mangled

Explanation: Destitute means without money,food and other things necessary for life

Q2. Orthodox (Give Antonym)

• A. Emangles
• B. Unconventional✓
• C. Conventional
• D. Traditional

Explanation: Orthodox means beliefs and behaviors that are generally accepted or approved (of a normal or usual type).

Q3. Economical (Antonym)

• A. Maddest
• B. Efficient
• C. Extravagant✓
• D. Expensive

Explanation: Economical means providing good service or value in proportion to the amount of money (for small amounts of money i.e.cheaply), resources or time.

Q4. Limitations( synonym)

• A. Shortcomings
• B. Restrictions✓
• C. Inability
• D. Vicinity

Explanation: Limitations means something which holds you back or limits you.

Q5. Expedient (synonym)

• A. Scapegoat
• B. Heterogeneous
• C. Suitable✓
• D. Apathetic

Explanation: Expedient means an action that is useful or convenient for a particular purpose, though not necessarily moral or fair.

Q6. Voracious:

• A. Reject
• B. Excitable
• C. Honest
• D. Greedy✓

Explanation: Voracious :wanting or devouring great quantities of food. The synonym for voracious is greedy

Q7. Hamper (antonym)

• A. Ogle
• B. Clarify
• C. Assist✓
• D. Uncover

Explanation: Hamper has two meanings. One meaning is a large basket used for carrying food,wine etc. The other meaning is to prevent somebody’s free movement or activity,to restrict or hinder something or somebody. Here the second meaning is being used.

Q8. What is the synonym of the following word:Monotonous:

• A. Limited
• B. Conclusive
• C. Extremist
• D. Repetitive✓

Explanation: The correct synonym for 'monotonous' is 'repetitive,' as both terms convey the idea of something that is dull, tedious, and lacking variety due to its repetitive nature. The other options are incorrect because they describe concepts unrelated to repetition: 'Limited' pertains to constraints, 'Conclusive' refers to decisiveness, and 'Extremist' relates to radicalism.

Q9. Intensive (antonym)

• A. Casual✓
• B. Aggressive
• C. Progressive
• D. Attentive

Explanation: Intensive means extremely thorough or concentrating all one’s efforts on a specific area.

Q10. Precise (synonym)

• A. Concise
• B. Comprehensive
• C. Exaggerate
• D. Accurate✓

Explanation: Precise means stated clearly and accurately.

Q11. In which of the following phylla, glass like silica is found?

• A. Mollusca
• B. Arthropoda
• C. Diatoms✓
• D. Starfish

Explanation: Diatoms are plant like protists (algae).The cell wall of each diatom consists of two shells that overlap where they fit together much like a petri dish.silica is deposited in the shell and this glass like material is laid down in intricate

Q12. They store food.

• A. Plastids✓
• B. Mitochondria
• C. Ribosomes
• D. Nucleus

Explanation: Plastids are classified into chloroplasts, leucoplasts and chromoplasts. LEUCOPLASTS act as storage organs.they are colourless and may be tubular triangular or some other shape ref ptb page 61.

Q13. Which of the following is replacd by thymine.

• A. Uracil✓
• B. Cytosine
• C. Guanine
• D. Adenine

Explanation: Cytosine guanine and adenine is present in both DNA and RNA .Cytosine is always equal to guanine and adenine is always equal to thymine {in DNA} and uracil {in RNA}

Q14. Which of the following is not an enzyme.

• A. Polymerase
• B. Hydrosere✓
• C. Isomerase
• D. Enzymate

Explanation: An enzyme is a biological catalyst.theyusly increase the efficiency of biological reaction

Q15. Which of the following is the most effective way of respiration in frog.

• A. Pulmonary
• B. Bucco-pharyngeal
• C. Both buccal and pharyngeal
• D. Cutaneous✓

Explanation: In an aquatic environment, cutaneous respiration is the most effective method of respiration for a frog. However, in a terrestrial setting, pulmonary respiration becomes more prominent. Both types work in combination, but cutaneous respiration is generally considered the most effective overall.

Q16. Plants convert solar energy to chemical energy present in the bonds of chemical substances namely:

• A. NADPH
• B. Nucleotide
• C. ATP✓
• D. RNA

Explanation: ATP is formed as a result of light reactions of photosynthesis . ATP provides chemical energy as a result of the breakage of the phosphate bond . Plants use the solar energy and convert it to chemical energy with the help of chloroplast through the process of photosynthesis.The chemical energy produced is stored in the bonds of chemical substances namely ATP. Ref ptb page 214.

Q17. The harmones which controls the contraction of uterus at the same time of birth is

• A. Prolactin
• B. Oxytocin✓
• C. Gonadotropin
• D. Antidiuretic hormone

Explanation: Oxytocin is produced by hypothalamus.Oxytocin’s release is stimulated by distension of cervix,decrease in progesterone level in blood,and neural stimuli during paturation and suckling.Primary action is on smooth muscles, particularly in the uterus during childbirth,and also causes milk ejection from mammary glands ref;ptb page74.

Q18. Substaneous fat accumulation is necessary for

• A. Seal
• B. Shark✓
• C. Elephant
• D. Kengaroo

Explanation: Sharks belong to class chondrichthyes. They are of great economic importance.With the exception of whales, sharks are the largest vertebrates. A layer of subcutaneous fat lies between the dermis and the underlying fascia. It helps to insulate the body from cold, cushions deep tissues from blunt trauma, and serves as a reserve source of energy for the body.

Q19. Hirudin anti clotting factor is found in

• A. Leech✓
• B. Earthworm
• C. Wuchereria
• D. Nereis

Explanation: Leeches produce an enzyme in their saliva called hirudin. Hirudin is a powerful anticoagulant. They produce this to prevent their host from forming a clot so that they can feast on blood more easily. When you apply a leech, the hirudin acts locally at the bite site to allow continued bleeding for 2-3 hours after the leech is removed. While attached, the leech actively removes some of the built up blood which eases the pressure within the tissue. After the leech is removed, the built up venous blood will continue to drain from the site where the leech was attached, which is therapeutic for the tissue. The effects of the hirudin may last up to a few hours. You can tell that it's no longer active when the blood stops draining. It is at that time that you should apply another leech, unless directed by the physicians to do otherwiseAnticoagulant Hirudin is extracted from leech which is used for medicinal purposes. It can remove the clots and is hence used extensively post surgeries.

Q20. Animals and plant characters are found in

• A. Euglena✓
• B. Paramecium
• C. Bacteria
• D. Viruses

Explanation: Euglenoids have been classified in the plant and animals kingdom . Based on the molecular data, euglenoids are closely related to zooflagellates. They are plant-like in their pigments. However, some photosynthetic euglenoids lose their pigments in dark and obtain their nutrients heterotrophically by ingesting organic matter. Other species of euglenoids are always colourless and heterotrophic. Ref ; ptb page 106.

Q21. The pathogenic bacteria transfer a DNA to another bacteria by

• A. Transduction
• B. Transformation
• C. Conjugation✓
• D. Fragmentation

Explanation: Conjugation is the direct transfer of DNA between two bacterial cells that are temporarily joined, often involving specialized structures like sex pili, and does not involve viruses.

Q22. When a prosthetic group joins with enzyme it is called:

• A. Cofactor
• B. Coenzyme
• C. Holoenzyme✓
• D. Apoenzyme

Explanation: A holoenzyme is formed by the combination of polypeptide chain and cofactor it is biochemically active ref page 37 ptb

Q23. What does not take place in meiosis

• A. Chromosomes multiplied✓
• B. Separation of chromosomes
• C. Production of hormones
• D. Production of enzymes

Explanation: Meiosis is a type of cell division in sexually reproducing organisms that reduces the number of chromosomes in gametes (the sex cells, or egg and sperm). In humans,Meiosis is a type of cell division that reduces the number of chromosomes in the parent cell by half and produces four gamete cells. This process is required to produce egg and sperm cells for sexual reproduction., body (or somatic) cells are diploid, containing two sets of chromosomes (one from each parent). REF PAGE 156 PTB

Q24. When phenotypically tall plants are crossed with pure dwarf plants what cross is

• A. Test Cross✓
• B. Monohybrid Cross
• C. Dihybrid Cross
• D. Multiple Cross

Explanation: MENDEL derived the test cross which is used to test the genotype of an individual showing a dominant phenotype .It is mating an individual showing dominant phenotype crossed with an individual showing a recessive phenotype . the cross finds whether the organism is homozygous or heterozygous dominant ref ptb page 167

Q25. In chick embryo the space between two layers is

• A. Coelom✓
• B. Blastocoel
• C. Primitive streak
• D. Gastrocoel

Explanation: The cavity formed between the somatic and splanchnic mesoderm is called coelom..REF PTB PAGE 112

Q26. In which of the following regions organic materials and shells are found

• A. Littoral
• B. Abyssal✓
• C. Euphotic
• D. Bathyal

Explanation: The seafloor and water column from 3,000 to 6,500 metres (9,842 to 21,325 feet) depth is known as the abyssal zone, or the abyss. Sunlight doesn't penetrate to these depths, so the waters here are extremely dark, and the animals that live here often use bioluminescence to communicate. Organic matter, organic material, or natural organic matter refers to the large source of carbon-based compounds found within natural and engineered,terrestrial, and aquatic environments. It is matter composed of organic compounds that have come from the faeces and remains of organisms such as plants and animals.

Q27. Which parasite can survive in artificial media?

• A. Endoparasites
• B. Ectoparasites
• C. Obligate parasites
• D. Facultative parasites✓

Explanation: - Facultative: This term refers to organisms, including parasites, that can adapt and survive in different environments, including artificial media. A facultative parasite is one that can live both as a parasite on a host organism and also independently in the environment. This ability to thrive in various conditions, including artificial laboratory media, makes facultative parasites versatile and adaptable.- Obligate: Obligate parasites are those that can only survive as parasites within a host organism. They are dependent on the host for their survival and cannot live independently outside of a host.- Ectoparasite: Ectoparasites are parasites that live on the surface of the host's body. While some ectoparasites might be able to survive on artificial media for a short period, they are primarily adapted to live on the host's body.- Endoparasite: Endoparasites are parasites that live inside the host's body. Similar to ectoparasites, some endoparasites might be able to survive on artificial media temporarily, but they are primarily adapted to thrive within the host's body.

Q28. In light antenna complex, is not found

• A. Haemoglobin✓
• B. Chlorophyll b
• C. Chlorophyll a
• D. Carotenoids

Explanation: The antennae complex contains many molecules of chlorophyll a, chlorophyll b and carotenoids most of them channelling the energy to the reaction centre.∙All of these pigments are generally embedded in the thylakoid membrane of the chloroplast present in the plant cells. Ref ptb page 215 antenna complex is an array of protein and chlorophyll molecules embedded in the thylakoid membrane of plants and cyanobacteria, which transfer light energy to one chlorophyll a molecule at the reaction centre. Haemoglobin (Hb) is a protein found in the red blood cells that carries oxygen in your body and gives blood its red colour. Haemoglobin levels vary from person to person. Men usually have higher levels than women.

Q29. If the concentration of CO2 and H+IONS increase in the body the rate of breathing will

• A. Increase✓
• B. Decrease
• C. Remain constant
• D. Voluntary

Explanation: If the level of CO2 increases the oxygen tension decreases and the capacity of haemoglobin to hold oxygen becomes less and hence the co2 tension favours the greater liberation of O2 from the blood to the tissue. If the H+ ions increase ,pH decreases and hydrogen ions combine with the protein part of the haemoglobin and decrease the oxygen binding ability of haemoglobin. So the increase in CO2 and H+ ions increase in the body the rate of breathing will increase

Q30. Lacterials require which muscles to flow

• A. Unstriated and involuntary✓
• B. Striated and voluntary
• C. Striated and involuntary
• D. Unstriated and voluntary

Explanation: As lacteals are made up of smooth muscles do they are non - striated and involuntary.

Q31. The theory of inheritance of acquired characters was a prposed by

• A. De Vries
• B. Darwin
• C. Lamarck✓
• D. Weismann

Explanation: Lamarck is best known for his Theory of Inheritance of Acquired Characteristics, first presented in 1801 (Darwin's first book dealing with natural selection was published in 1859): If an organism changes during life in order to adapt to its environment, those changes are passed on to its offspring.ref ptb page223.

Q32. Which is associated with the long term memory

• A. Hippocampus✓
• B. Hypothalamus
• C. Brain stem
• D. Pons

Explanation: Hippocampus is a complex brain structure embedded deep into the temporal lobe.HIPPOCAMPUS plays a very important role in the formation of long term memory and is required for learning. Ref ptb page 69

Q33. Pituitary gland is controlled by which of the following

• A. Adrenal cortex
• B. ACTH
• C. Hypothalamus✓
• D. Cytoxan

Explanation: It is a part of the forebrain.It is here that many of the sensory stimuli of the nervous system are converted into hormonal responses.it is believed that oxytocin and ADH are produced by hypothalamus and released by the pituitary. The hypothalamus secretions control the secretion of the pituitary gland. ref ptb page 72.

Q34. Bacteria takes DNA in the presence of

• A. NaOH
• B. HCl
• C. Calcium chloride✓
• D. NaCl

Explanation: Bacteria can take up foreign DNA in a process called transformation. Transformation is a key step in DNA cloning. It occurs after restriction of digestion and ligation and transfers newly made plasmids to bacteria. After transformation, bacteria are selected on antibiotic plates. Transformation takes place in the presence of CALCIUM CARBONATE. Ref ptb page 133

Q35. Centromeres at the end of chromosomes

• A. Acrocentric
• B. Metacentric
• C. Submetacentric
• D. None of the above✓

Explanation: Explantion is given below:

Q36. In plant cell, cell wall and cell plates is the product of

• A. Golgi apparatus✓
• B. Mitochondria
• C. Endoplasmic Reticulum (ER)
• D. Plastids

Explanation: At cytokinesis, in place of the contractile ring a membrane structure,phragmoplast is formed from minute vesicles produced in the Golgi apparatus present in the cytoplasm.the vesicles originate during the metaphase line up in the centre of the dividing cell where they fuse to form the phragmoplast at the end of telophase.the membrane of vesicles forms the plasma membrane of the daughter cells. The vehicles also contain materials for future cell wall ref ptb page 154.

Q37. Tissue fluid do not contain

• A. Wbcs
• B. Glucose
• C. Amino acids
• D. Fibrinogen✓

Explanation: The pressure within the capillaries causes a continuous leakage of the fluid from the blood into the spaces that surround the capillaries and tissues.the fluid known as the interstitial fluid Consists primarily of water ,in which dissolved nutrients.hormones,gases,wastes, glucose ,WBCs and small proteins are present .Tissue fluid is important for the exchange of substances between the blood and the cells.fibrinogen is a large protein and cannot squeeze through the holes of the capillary wall. Ref ptb page 318.

Q38. In which of the following cycle of plasmodium which of the following is not affected

• A. Sporozoite
• B. Merozoite
• C. Cryptozoite✓
• D. Trophozoite

Explanation: During a blood meal, a malaria-infected female Anopheles mosquito inoculates sporozoites into the human host . Sporozoites infect liver cells and mature into schizonts , which rupture and release merozoites . (Of note, in P. vivax and P. ovale a dormant stage [hypnozoites] can persist in the liver (if untreated) and cause relapses by invading the bloodstream weeks, or even years later.) After this initial replication in the liver (exo-erythrocytic schizogony ), the parasites undergo asexual multiplication in the erythrocytes (erythrocytic schizogony ). Merozoites infect red blood cells . The ring stage trophozoites mature into schizonts, which rupture releasing merozoites . Some parasites differentiate into sexual erythrocytic stages (gametocytes) . Blood stage parasites are responsible for the clinical manifestations of the disease. The gametocytes, male (microgametocytes) and female (macrogametocytes), are ingested by an Anopheles mosquito during a blood meal . The parasites’ multiplication in the mosquito is known as the sporogonic cycle . While in the mosquito’s stomach, the microgametes penetrate the macrogametes generating zygotes . The zygotes in turn become motile and elongated (ookinetes) which invade the midgut wall of the mosquito where they develop into oocysts . The oocysts grow, rupture, and release sporozoites, which make their way to the mosquito’s salivary glands. Inoculation of the sporozoites into a new human host perpetuates the malaria life cycle.

Q39. Which enzyme is used to cut dna into smaller pieces

• A. DNA polymerase
• B. Vector
• C. Restriction enzymes✓
• D. DNA ligase

Explanation: these are natural enzymes , which a bacteria uses against the attack of viruses A restriction enzyme is a protein isolated from bacteria that cleaves DNA sequences at sequence-specific sites, producing DNA fragments with a known sequence at each end. The use of restriction enzymes is critical to certain laboratory methods, including recombinant DNA technology and genetic engineering REF PTB PAGE 203

Q40. Number of chromosomes in human are

• A. 46✓
• B. 48
• C. 44
• D. 56

Explanation: A human cell has 23 pairs of chromosomes hence 46 chromosomes in total

Q41. Grignard reagent on reaction with methanol gives

• A. Primary alcohol
• B. Secondary alcohol
• C. Ethane✓
• D. Tertiary alcohol

Explanation: Grignard reagents are organometallic species with formula R-Mg-X. The reaction of Grignard reagents with alcohols is a type of acid – base reaction. The reaction involves the acidic nature of the O – H bond of the alcohols, as the O – H bond has removal of hydrogen.

Q42. A compound X oxidises in the presence of KMNO4 of concentrated H2SO4 to form a compound Y, which one's reaction with X gives a sweet/pleasant smell?

• A. Aldehyde
• B. Ester✓
• C. Ether
• D. Acetone

Explanation: Most ALDEHYDES have a very strong scent with a slightly greasy waxy character with a thin almost metallic twist. Other aldehydes can also give off a citrus-like, mandarin-like fragrance or a rose-like accord.

Q43. The total pressure of the gas is equal to the sum of the partial pressure of all the individual gasses due to the foloowing

• A. The total pressure results from the collision of gas molecules with the walls of the container.✓
• B. The total pressure arises from the effective collision of all individual gas molecules with each other.
• C. The total pressure comes from the collision of gas molecules with the bottom of the container.
• D. None of these options describe the reason for the total pressure.

Explanation: The partial pressure of a gas in a mixture of gases is the pressure that it would exert on the walls of the container, if it were present all alone in that same volume under the same temperature.According to Dalton's law the total pressure exerted by a mixture of non-reacting gases is equal to the sum of their individual partial pressures.’ Let the gases are designated as 1,2,3, and their partial pressures are p1 , p2 , p3 .The total pressure (P) of the mixture of gases is given by Pt = p1 + p2 + p3 REF PTB PAGE 22.

Q44. Metallic character of AL3+ is greater than boron

• A. Due to less electron population
• B. The number of neibouring atoms in lactic increase
• C. Both of the above✓
• D. None of the above

Explanation: The maximum capacity of electron accommodation of boron in its outermost shell is 8 electrons and that of aluminium is 18 electrons. It means the electron population of aluminium is less than boron. Due to less electron population, the number of neighbouring atoms in the latest increases. That is why metallic character of aluminium is greater than boron.

Q45. NaNO3 +heat->NaNO2+O2

• A. Kp > Kc✓
• B. Kp < Kc
• C. Kc = Kp
• D. Kp = Ksp

Explanation: Due to the difference in moles kp is not equal to Kc they are related by the following relation and the moles of product are greater than reactant hence Kp>Kc

Q46. Following are not metal of group 2A

• A. Mg Be Sr Ra
• B. Be Ra Rb Ba✓
• C. Mg Be Ca Sr
• D. None

Explanation: HENCE group 2A INCLUDES BE Mg Ca Sr Ba Ra whereas Rb belongs to GROUP 1A SO B IS CORRECT

Q47. When the reaction is in equilibrium

• A. The concentration of products is equal to the concentration of reactants
• B. The rate of the forward reaction is equal to the rate of the backward reaction✓
• C. The concentration of products is not equal to the concentration of reactants
• D. All of the above

Explanation: The concentration of reactanrs and products at equilibrium depends on value of KC

Q48. Hydrogen has ground state energy of

• A. 1.5 eV
• B. 13.6 eV✓
• C. -0.8 eV
• D. 23.3 eV

Explanation: The ground state of a quantum-mechanical system is its stationary state of lowest energy; the energy of the ground state is known as the zero-point energy of the system. An excited state is any state with energy greater than the ground state.ground state energy of hydrogen atom is E=−13.67eV<fact>The Kinetic energy is 13.67 ev and potential energy is 27.2 ev

Q49. Hydrogen is not absorbed by

• A. Co✓
• B. Ni
• C. Pt
• D. Pd

Explanation: Hydrogen absorption is electronically an incorporation of electrons and protons into the electronic structure of the host lattice. The protons lower the energy of a part of the d-electrons and they lead to new states about 4 eV below the Fermi energy EF.

Q50. Bond energy ID need to know

• A. Nature of compound✓
• B. Nature of ionic compound
• C. Nature of bonding
• D. Nature of reaction

Explanation: Bond energy serves a very important purpose in describing the structure and characteristics of a molecule. It can be used to determine which Lewis Dot Structure is most suitable when there are multiple Lewis Dot Structures.Bond energy is the measure of bond strength in a chemical bond. It is the energy or heat required to break one mole of molecules into their individual atoms.Bond energy is based on an average of bond dissociation values for species in the gas phase, typically at a temperature of 298 Kelvin. It may be found by measuring or calculating the enthalpy change of breaking a molecule into its component atoms and ions and dividing the value by the number of bonds

Q51. 1 ounce pressure of a gas at sip contains

• A. 224000 cubic meter
• B. 22.4 cubic meter
• C. 22.4000 cubic meter
• D. 0.00224 cubic meter✓

Explanation: 1 OUNCE PRESSURE OF A GAS AT STP COVERS 0.00224 cubic meter. volume the remaining options do not show correct numerals . The volume of a gas can be determined by the following formula PV = nRT

Q52. Swollen eyes , tongue and mouth is caused by

• A. Vitamin B1 (Thiamine)
• B. Vitamin B2 (Riboflavin)✓
• C. Vitamin B12 (Cobalamin)
• D. Vitamin B6 (Pyridoxine)

Explanation: The signs and symptoms of riboflavin deficiency (also known as ariboflavinosis) include skin disorders, hyperemia (excess blood) and edema of the mouth and throat, angular stomatitis (lesions at the corners of the mouth), cheilosis (swollen, cracked lips), hair loss, reproductive problems, sore throat, itchy and red.

Q53. 66 Nylon is a polymer of

• A. Vinyl Chloride
• B. Aldehyde
• C. Adipic acid and hexamethylenediamine✓
• D. Both A and B

Explanation: Vinyl chloride polymerizes to form poly vinyl chloride Aldehydes have a low ceiling temperature (Tc) which is either at or noticeably below ambient temperature which means that these compounds do not polymerize at or above room temperature.These resins are formed by the condensation of polyamines with aliphatic dicarboxylic acids.One of the most famous condensation polymers discovered is Nylon. The word Nylon has been accepted as a generic name for synthetic polyamides. Nylon 6, 6 is the most important polyamide. It is obtained by heating adipic acid (hexanedioic acid) with hexamethylenediamine. Nylon 6,6 derives its name from its starting materials adipic acid and hexamethylenediamine,both of which have six carbon atoms. Nylon is mainly used as a textile fibre. It has a combination of high strength, elasticity, toughness and abrasion resistance REF PTB.

Q54. The traid contains CL 35.5 and iodine 121 gives the atomic mass of bromine

• A. 80
• B. 81✓
• C. 72
• D. 90

Explanation: Dobereiner's triads- It states that the atomic mass of the middle element in doberiener's triad is exactly or nearly the arithmetic mean of the atomic masses of other two elements.

Q55. Anode mud, in blistening of copper is

• A. Impure left✓
• B. Pure Cu
• C. Electrolyte
• D. Blistening copper

Explanation: Impure copper is refined by the electrolytic method in the electrolytic cell. Impure copper acts as anode and a pure copper plate acts as cathode.Copper sulphate solution in water is used as an electrolyte. Oxidation reaction takes place at the anode. Copper atoms from the impure copper lose electrons to the anode and dissolve in solution as copper ions: Reduction reaction takes place at the cathode. The copper ions present in the solution are attracted to the cathode.Pure copper is deposited on the cathode. Where they gain electrons from the cathode and become neutral and deposit on the cathode. In the process, impure copper is eaten away and purified copper atoms deposit on the cathode . the impurities settle down as anode mud

Q56. Reforming is the process to convert

• A. Smaller compounds into larger ones
• B. Straight-chain hydrocarbons into branched-chain hydrocarbons✓
• C. Less volatile compounds into more volatile ones
• D. Branched-chain hydrocarbons into cyclic compounds

Explanation: The gasoline fraction present in petroleum is generally not of good quality. When it burns in an automobile engine, combustion can be initiated before the spark plug fires. This produces a sharp metallic sound called knocking which greatly reduces the efficiency of an engine. The quality of a fuel is indicated by its octane number. As the octane number increases, the engine is less likely to produce knocking. Straight- chain hydrocarbons have low octane numbers and make poor fuels. Experiments have shown that isooctane or 2,2,4- trimethyl pentane burns very smoothly in an engine and has been arbitrarily given an octane number of 100. The octane number of gasoline is improved by a process called reforming. It involves the conversion of straight chain hydrocarbons into branched chains by heating in the absence of oxygen and in the presence of a catalyst. REF PTB CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2—>CH3-C(CH3)2-CH2-CH(CH3)-CH3 THE REACTION TAKES PLACE IN THE PRESENCE OF A CATALYST

Q57. In destructive distilation of coal oil the following are obtained except

• A. Coal liquor
• B. Coal tar
• C. Colle✓
• D. Coal gas

Explanation: The main products formed after the destructive distillation of coal include coke, ammoniacal liquor (or ammonium hydroxide) coal gas and coal tar. Coal tar < Ammonia < Coal gas <coke

Q58. Azimuthal quantum number show

• A. Orientation in space
• B. Sizes of orbital
• C. Shape of orbital✓
• D. None

Explanation: Explanation is given below:

Q59. In ethane. Which of the following is correct

• A. Only carbon is hybridized
• B. No hybridization occurs in ethane
• C. Carbon atoms are sp3 hybridized in ethane✓
• D. Carbon atoms in ethane are sp2 hybridized

Explanation: Atomic orbitals differing slightly in energy intermix to form new orbitals, which are called hybrid atomic orbitals. They differ from the parent atomic orbitals in shape and possess specific geometry. The atomic orbital hybridization gives a satisfactory explanation for the valency of the elements. In some cases, the electrons belonging to the ground state are promoted to the excited state as a result of which there is an increase in the number of unpaired electrons. The electronic distribution of carbon atom should be kept in mind to understand intermixing of orbitals in C2H6, 1 s orbital and three p-orbitals (px, py, pz) take part in hybridization. There is a formation of four sp3 hybridised orbitals. During the hybridization of ethane four identical bonds are formed in a perfect

Q60. In SN2 reaction, The rate is depending on

• A. Nu
• B. Substrate
• C. Eu
• D. Both A and B✓

Explanation: Nucleophilic substitution reactions on alkyl halides involve two main processes, the breakage of C — X bond and the formation of C — Nu bond. The mechanism of the nucleophilic substitution reactions depends upon the timing of these two processes. If the two processes occur simultaneously the mechanism is called SN2 . If the bond breaks first followed by the formation of a new bond, the mechanism is called SN1. Kinetic studies of the reactions involving SN2 mechanism have shown that the rates of such reactions depend upon the concentrations of alkyl halide as well as the attacking nucleophile. Mathematically, the rate can be expressed as: Rate = k [Alkyl halide]1 [Nucleophile]1

Q61. Unsaturated compounds are formed by the following:

• A. Elimination reaction✓
• B. Substitution reaction
• C. Nucleophilic substitution
• D. Electrophilic reaction

Explanation: Unsaturated compounds are chemical compounds which contain one or more carbon-carbon multiple bonds i.e. double or triple bonds such as those found in alkenes and alkynes respectively. Since double and triple bonds contain pi-bond, unsaturated compounds can also be defined as compounds having carbon-carbon pi-bonds.

Q62. Glucose and fructose are linked by:

• A. 2,4-glycosidic linkage
• B. 1,4-glycosidic linkage
• C. 1,2-glycosidic linkage✓
• D. 1,1-glycosidic linkage

Explanation: Alpha-glucose and beta-fructose are linked by alpha-1,2-glycosidic linkage in sucrose. The sucrose is formed by the condensation of glucose and fructose. In this reaction. The –OH group at C-1 of glucose interacts with the –OH group at C-2 of fructose, liberating a water molecule forming alpha-1,2-glycosidic linkage.

Q63. Compound having different crystalline forms this statement is related to:

• A. Allotropy
• B. Isomerism
• C. Polymorphism✓
• D. Isomorphism

Explanation: The phenomenon where a compound exists in more than one crystalline form is called polymorphism.Example: NaCl is found in cubic and octahedral forms. HgI2 is found in orthorhombic and tetragonal forms

Q64. Products of exothermic reaction are

• A. Unstable
• B. Varies between stable and unstable
• C. Stable✓
• D. None of the above

Explanation: For an exothermic reaction, the products have less potential energy than the reactants and so they are more stable than the reactants because the lower the potential energy of a molecule, the more stable it is. On the other hand, for an endothermic reaction the products have more potential energy than the reactants so they are unstable compared to the reactants.

Q65. Rutherford discovered that the nucleus of an atom has charge

• A. No charge
• B. Positive✓
• C. Negative
• D. Very less

Explanation: Since a nucleus is made up of neutrons and protons and neutrons carry no charge and protons have positive charge,so the nucleus of an atom is positive.

Q66. The heat absorbed at constant volume is equal to:

• A. External pressure
• B. Energy change✓
• C. Internal pressure
• D. All of the above

Explanation: Explanation: According to the first law of thermodynamics: △E= q-P△V (1)where E is the internal energyq is the heat absorbed by the systemP△V is the work done on the systemIf volume is constant then △V=0 and so the term P△V becomes 0.Now for constant volume system equation 1 can be written as: △E=qvSo the heat absorbed by the system at constant volume changes its internal energy. It does not affect the internal or external pressure.Thus option B change in energy is the correct answer.

Q67. Equilibrium is established when

• A. The rate of forward reaction = the rate of reverse reaction✓
• B. The rate of forward reaction > the rate of reverse reaction
• C. The rate of forward reaction < the rate of reverse reaction
• D. The rate of forward reaction rate of reverse reaction

Explanation: Equilibrium is the state of a reversible reaction where the rate of forward reaction is equal to the rate of reverse reaction.

Q68. Rate of the reaction depends on the following factors except:

• A. Concentration
• B. Temperature
• C. Pressure
• D. Molecularity✓

Explanation: Molecularity of a reaction is the number of molecules of reactants participating in the reaction. It does not affect the rate of reaction.So this is the correct answer.

Q69. The electronic configuration of a noble gas is:

• A. ns2 np6✓
• B. ns2 np6 nd8
• C. ns2 np8
• D. None of the above

Explanation: Since noble gases have 8 valence electrons, they are present in group VIII. For group VIII-A the electronic configuration is ns2 np6.

Q70. Cathode rays do not depend upon:

• A. Neither on gas nor on the material of the cathode✓
• B. Neither on H2SO4 nor on the liquid used
• C. All of the above
• D. None of the above

Explanation: Since cathode rays are electrons and electrons are the same for all matter, that’s why the nature of cathode rays does not depend on the gas used in the glass tube or the material of the cathode.

Q71. If T of a string is increased four times what will be the speed of wave:

• A. v/2
• B. v/4
• C. 2v✓
• D. 4v

Explanation: Explanation: The wave speed on string is: v =T/µWhere v is the velocity of the waveT is the tension of the string is the linear density of the string. ( Linear density is mass per unit length).So the wave speed is directly proportional to the square root of tension of the string. v∝TSo if tension quadruples, velocity doubles.

Q72. Sone is the unit of loudness. What is the relationship of sone with decibels?

• A. 1 sone = 40 dB
• B. 1 sone = 40 dB at 100 Hz
• C. 1 sone = 40 dB at 1000 Hz✓
• D. 1 sone = 40 dB at 10,000 Hz

Explanation: According to Steven’s definition, a loudness of 1 sone is exactly equal to 40 db at 1000 Hz (1kHz).

Q73. Three sources each of 1.5 volt are connected in series such that the negative terminal of one battery is connected with the positive terminal of the other battery.what is the net emf:

• A. 3 volt
• B. 4.5 volt✓
• C. 15 volt
• D. 6 volt

Explanation: When voltage sources are connected in the series facing the same direction, their voltages are added algebraically to get net emf. As all the cells are connceted end to end in such away that the lower potential of one battery is connected to the higher potential of other battery,voltage of every battery will be taken positive.So 1.5+1.5+1.5=4.5Hence, B is the correct answer.

Q74. Electromagnetic waves depend upon:

• A. Permittivity of the medium only
• B. Permeability of the medium only
• C. Both permittivity and permeability of the medium✓
• D. Nature of the medium

Explanation: Explanation: The speed of electromagnetic wave in a medium is given by: V=1/µ∊Where is the permeability of the medium and is the permittivity of the medium. This means the electromagnetic wave depends on the electric and magnetic properties of the medium which are defined by its permittivity and permeability.

Q75. If light enters from air to denser medium of “n” refractive index, then wave speed:

• A. Increases “n” times
• B. Decreases “n” times✓
• C. Decreases
• D. Remains same

Explanation: The refractive index is the measure of the bending of a ray of light when passing from one medium into another.It is also a measure of how much light slows down in any medium.Mathematically, it is the ratio of the velocity of light in a vacuum to its velocity in a specified medium i.e. n=c/v v=c/nwhere n is the refractive index c is the speed of light in vacuum or air v is the speed of light in any specified mediumSo from the above equation, it is clearly evident if light enters a medium of “n” refractive index from air, its speed decreases by “n”.

Q76. Shadow is formed due to the phenomenon of:

• A. Interference
• B. Polarisation
• C. Diffraction
• D. Rectilinear Propagation of Light✓

Explanation: A shadow is an interception of light.The phenomenon of rectilinear light propagation causes shadows to form. When a person, animal, or object obstructs a portion of a light source, a shadow is produced.Shadows form in the opposite direction of whatever is blocking that source.

Q77. If A.B=0, then between them is:

• A. 0
• B. 45
• C. 60
• D. 90✓

Explanation: Dot product of two vectors is obtained by multiplying their magnitudes and the cos of angle between them. It is orepresented by a dot(.) between the two vectors. A.B=ABcosNow if A and B are nonzero vectors, their dot product can only be zero if cos is zero. As cos 90=0, hence the angle between them is 90 degrees.BEWARE: cos 0=1

Q78. Energy of hydrogen atom in ground state is:

• A. -13.6eV✓
• B. -0.65eV
• C. -10.2eV
• D. -0.25eV

Explanation: Explanation: Expression for the energy of th nth allowed orbit in the hydrogen atom is: En=- 13.6eVn In ground state, n=1So E=-13.6eV

Q79. Dot product of force and velocity is:

• A. Rate of energy
• B. Rate of work✓
• C. Rate of power
• D. Rate of change of momentum

Explanation: Dot product of two vectors always gives a scalar. Dot product of force and velocity is power or rate of work.

Q80. If ʎ is decay constant then half life is:

• A. 0.693
• B. 0.693/ʎ✓
• C. ʎ/0.693
• D. 0.963/ʎ

Explanation: The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ.

Q81. When a body moves downward on inclined path in the presence of friction “f”, force on the body is:

• A. W sinØ - f✓
• B. W cosØ - f
• C. W cosØ
• D. F - W sinØ

Explanation: Explanation: The main forces on the body moving down an inclined path are friction and weight.On an inclined plane, the object moves down due the horizontal component of weight(mgsin or Wsin). If there is friction then net force on the object will be WsinØ-f.Fnet= WsinØ- f

Q82. According to thermodynamics, at constant volume Q=?

• A. Q=0
• B. Q=△W
• C. Q=△E✓
• D. Q=△E+P△V

Explanation: A process which occurs at constant volume is called an isochoric process.According to the first law of thermodynamics: △E= q-P△ V (1)where E is the internal energy q is the heat absorbed by the system P△V is the work done on the systemIf volume is constant then△ V=0 and so the term P△V becomes 0.Now for constant volume system equation(1)can be written as: △ E=qvSo the heat absorbed by the system at constant volume changes its internal energy.

Q83. Velocity of sound is:

• A. Directly proportional to the absolute temperature
• B. Directly proportional to the square root of absolute temperature✓
• C. Directly proportional to the pressure
• D. Directly proportional to the square root of centigrade temperature

Explanation: The velocity of sound in gas is:So velocity of sound depends only on the absolute temperature of the gas, molar mass of gas and gamma( which is specific for every gas).

Q84. The velocity of alpha particles is:

• A. 0.2 * 10^8 m/s
• B. 0.5 * 10^8 m/s
• C. 0.3 * 10^8 m/s
• D. 0.1 * 10^8 m/s✓

Explanation: Alpha particle shoots out from the nucleus with a high velocity of 0.1 *10^8 m/s.STB Pg-331.

Q85. Question is given below:

• A. 0✓
• B. 1
• C. -1
• D. 1/2

Explanation: The cross product of two parallel vectors is always zero because the sine of the angle between them is zero. In this case, i × i = |i||i| sin(0) = 0. This means that when dealing with parallel vectors, the resulting cross product will always be zero, as seen in Option A. Options B, C, and D are incorrect because they do not consider the fact that parallel vectors have a sine of 0, resulting in a cross product of 0.

Q86. The carrier waves are over modulated to be distorted version of modulating signal when:

• A. m=1
• B. m>1✓
• C. m<1
• D. m=1

Explanation: Amplitude modulation is the modification of the carrier's amplitude in response to the modulating signal.When the modulation is so strong that the carrier wave clips, the signal is said to be excessively modulated (distorts in AM).The modulation index of an amplitude modulated signal is defined as the measure or extent of amplitude variation about an un-modulated carrier. This can be expressed in mathematical terms as below:Modulation Index,m m=M/AWhere: A = the carrier amplitude. M = the modulation amplitude .If this m>1, overmodulation takes place and the carrier signal is distorted. Over-modulation is the result of raising the level of modulation above a modulation index of 1, or more than 100% modulation.

Q87. If the axis of rotation passes through the body and the body moves around then the motion is called:

• A. Spin motion✓
• B. Orbital motion
• C. Rectilinear motion
• D. Translatory motion

Explanation: Spin motion is the motion of an object as it rotates around an axis through its center of mass. Example: rotational motion of earth

Q88. Which instrument is used to detect the path of visible rays:

• A. Geiger counter
• B. Wilson cloud chamber
• C. Solid state detector
• D. None of the above✓

Explanation: None of the above devices is used for detecting the path of visible radiation.

Q89. Number of lines originating from q to the surface area of sphere is:

• A. Flux intensity
• B. Flux density✓
• C. Flux linkage
• D. Flux region

Explanation: Number of lines originating from the charge is the electric field. Electric field density is electric flux per unit area held perpendicular to electric field lines. So it is an electric field.

Q90. Relation between linear velocity and angular velocity is

• A. v = rω✓
• B. r = vω
• C. s = r
• D. v = △s/△t

Explanation: Let Δθ be the angular displacement of the body.Let "Δt" be the time taken.Then angular velocity of the body is ω=△s/△t (1)Now let v be the linear velocity of the body,then v=△S/△t (2)where △S is the arc length covered by the body in △t.As we know; △S=r△θSubstituting in (2), we get v=r△θ/△t v=r△θ/△t From (1), we get v=rω

Q91. Gauss’s law can be used to find electric field intensity:

• A. Charge enclosed in a sphere
• B. Charge placed between two oppositely charged surfaces
• C. Charge on an infinite sheet of paper
• D. All of the above✓

Explanation: Gauss’s laws states that the net electric flux through any closed surface is 1/o times the total charge enclosed in it. O=Q/∊oIt can be used for calculating the electric field in the case of charge distribution by considering a gaussian surface and applying gauss’s law. It can be used to find the electric field of an infinite sheet of paper, two oppositely charged plates and charge placed inside a sphere.

Q92. Instrument having very high magnification used to see microorganisms clearly

• A. Simple microscope
• B. Compound microscope✓
• C. Telescope
• D. Magnifying glass

Explanation: A microscope is a tool used to see items that are too tiny to be seen by the human eye. The most popular kind of microscopes are light microscopes, often called optical microscopes. Images are magnified by optical microscopes using light and lenses.

Q93. If radius of earth decrease by half and mass remains unchanged, then the value of “g” is:

• A. 4g✓
• B. 2g
• C. g/4
• D. g/2

Explanation: the formula for gravitational constant “g” is: g=GM/R2 (1)Where G is universal gravitational constant M is mass of the earth R is the radius of the earth.The new value of "g" is now obtained from equation(1) by substituting R'=R2as follows: g'=GM/R2/4 g'=4GM/R2 g'=4gSo new gravitational acceleration g' is four times its initial value.The right response is thus option A, which states that if the earth's radius were cut in half but its mass remained constant, the acceleration caused by gravity would increase by four times.

Q94. Ratio of emf in a potentiometer is directly proportional to ratio of:

• A. Balancing lengths✓
• B. Balancing widths
• C. Balancing heights
• D. None of the above.

Explanation: A potentiometer is a null type resistance network device for measuring potential differences. Balancing length is the length for which the galvanometer does show deflection because the current passing through it is zero. For potentiometer: ∊1/∊2=i1/i2Where l1 and l2 are lengths at which e.m.f 1 and 2 are balanced respectively. So they are balancing lengths. So for a potentiometer the ratio of emf is equal to the ratio of balancing lengths.

Q95. What is the formula of concave lens without convention of sign:

• A. 1/q - 1/q = 1/f
• B. 1/p + 1/q = 1/f✓
• C. 1/p - 1/p = 1/f
• D. 1/p - 1/q = 1/f

Explanation: A concave lens is a diverging lens. Without sign convention B is the correct answer. But with sign convention D is correct because is forms a virtual image.

Q96. If the power of a lens in 2 diopter and of the other is -4 diopter what is the focal length of the combination of lens:

• A. 20cm
• B. 50cm✓
• C. 100cm
• D. 500cm

Explanation: The power of combination of lenses is the algebraic sum of the individual powers of the lenses.So P= +2-4= -2 diopterNow focal length is: f=1P f=½= o.5m= 0.5100= 50 cm

Q97. In a solid state conductor:

• A. Sealed tube is used
• B. Argon is used
• C. P-N junction is used✓
• D. All of the above can be used

Explanation: Solid state detectors are also called semiconductor radiation detector.Radiation detectors that use a semiconductor material such as silicon or germanium as the detecting medium are known as solid-state semiconductor detectors. When these detectors come into contact with ionizing radiation, they generate a pulse of electric current using pairs of charge carriers, electrons and holes.

Q98. Light having range of 7.610-7cm to 4.010-7 cm lies in the region

• A. Infrared radiation
• B. X-rays
• C. Visible radiation✓
• D. U.V radiation

Explanation: The electromagnetic spectrum categorizes electromagnetic radiation based on wavelength. The range of 7.610-7 cm to 4.010-7 cm corresponds to approximately 760 nm to 400 nm, which is the range for visible light, making option C the correct choice. Options A and D are incorrect because they refer to regions with longer (Infrared) and shorter (UV) wavelengths, respectively, while option B is incorrect as X-rays have much shorter wavelengths than those given in the question.

Q99. Which is the type of collision in which both the linear momentum and the kinetic energy of the system remain conserved?

• A. Inelastic Collision
• B. Elastic Collision✓
• C. Both a and b
• D. None of the options

Explanation: In an elastic collision, the linear momentum and the kinetic energy of the system remains conserved.